# In $ZF$, $AC$ is equivalent to $\forall \alpha (\mathscr P(\alpha)$ can be well-ordered) [duplicate]

This is an exercise from Kunen - An introduction to independence proofs that I have hard time to solve.

In $ZF$, $AC$ is equivalent to $\forall \alpha (\mathscr P(\alpha)$ can be well-ordered)

## marked as duplicate by Asaf Karagila♦ axiom-of-choice StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 9 '15 at 19:23
• This is not the same question, your answer claimed that "$\forall X$ if $X$ is well-orderable then so is $2^X$" implies AC. – nombre Dec 9 '15 at 22:09
• @mathcounterexamples.net: AC implies the other result since it implies Zermelo's theorem which is $\forall \alpha$, $\alpha$ can be well-ordered. As for the other implication, if there is no restriction on $\alpha$ (if ou don't want it to be an ordinal) you can start by proving that if $\alpha$ embeds in a well-orderable set then it is well-orderable, then use Zermelo $\longrightarrow$ AC. In any case, the difficult result is the equivalency of Zermelo's theorem and AC. – nombre Dec 9 '15 at 22:16